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A point is moving along the curve $y = x^2$ with unit speed. What is the magnitude of its acceleration at the point $(1/2, 1/4)$?

My approach : I use the parametric equation $s(t) = (ct, c^2t^2)$, then $v(t) = s'(t) = (c, 2c^2t)$ and $a(t) = v'(t) = (0, 2c^2)$. Now the point $(1/2, 1/4)$ is reached at time $t = \frac{1}{2c}$, so $v(\frac{1}{2c}) = (c, c)$. Now the unit speed condition gives us $\sqrt{c^2 + c^2} = 1 \implies c = \frac{1}{\sqrt{2}}$. So the magnitude of acceleration is $2c^2 = 1$.

But the answer is $\frac{1}{\sqrt{2}}$. Can someone help me what is wrong in my approach.

walter
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2 Answers2

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What is wrong is that using the parametric coordinates

$$s(t)=(ct,c^2t^2)$$

the point is not moving with unit speed. This can be easily seen by computing the norm of the speed, which is not constant. You need to normalize the speed for all $t$. Then find the time at which the point $(1/4,1/2)$ is reached. And finally compute the acceleration at this point.

mathcounterexamples.net
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  • Oh, I see. Thank you a lot for pointing that out. Also, can you please help me how the parametric equation should look so that the speed is unit for all $t$. – walter Aug 09 '21 at 14:50
  • To get the right parametric equation, you have to divide the speed you currently have by its norm. – mathcounterexamples.net Aug 09 '21 at 14:56
  • Don't think it's easy compute. Assuming $c=1$ the normalized velocity vector becomes $\frac {(1,2t)} {\sqrt {1 + 4t^2}}.$ I have to find the displacement vector now. So the integrating factors should be such that the displacement along the $y$-axis is the square of the displacement along the $x$-axis. Then we have to determine the time $t_0$ when the displacement vector is at the given point. Differentiating the displacement vector twice and evaluate it at $t_0$ would give the acceleration vector and finally we need to find it's magnitude or norm. It's not that easy as you are anticipating. – Anacardium Dec 30 '23 at 22:29
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You can use Cartesian coordinates for this. You have $$y=x^2\implies \dot{y}=2x\dot{x}$$ Therefore at $x=\frac12$, $\dot{y}=\dot{x}$

For constant speed we have $${\dot{x}}^2+{\dot{y}}^2=1$$ Hence at $x=\frac12$, $\dot{y}=\dot{x}=\frac{1}{\sqrt{2}}$

Differentiating with respect to time the curve again, we have $$\ddot{y}=2{\dot{x}}^2+2x\ddot{x}$$

Therefore at $x=\frac12$, we have $$\ddot{y}=1+\ddot{x}$$

Finally, differentiating the constant speed condition gives $$2\dot{x}\ddot{x}+2\dot{y}\ddot{y}=0\implies \ddot{x}=-\ddot{y}$$

So $\ddot{y}=\frac12$ and $\ddot{x}=-\frac12$ and hence the result.

David Quinn
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